ESF02: Coulomb's Law in Advanced Coordinate Systems

[@bignick26]

At the time 1:16:42 when substituting the r's into the integral, where does the (r') go in the numerator? The way I understood it, r = xX^ + yY^ + 0Z^, and (r-r') = xX^ + yY^ - x'X^ - y'Y^ - z'Z^. Therefore the numerator which I thought was (r')(r-r'), would be different than simply (r-r'). Am I missing something here?

[@bignick26]

I may have answered my own question. In earlier equations for E(r), the (r') was not present in the numerator, so it may either be a functional variable of rho or a misprint perhaps?

[@Kashif_Javaid]

Which book is referenced in this lecture?

[@gokhansatlms]

holographic blackboard is needed for also watcher :)